Problem: Multiply the following complex numbers: $({4+5i}) \cdot ({2+3i})$
Solution: Complex numbers are multiplied like any two binomials. First use the distributive property: $ ({4+5i}) \cdot ({2+3i}) = $ $ ({4} \cdot {2}) + ({4} \cdot {3}i) + ({5}i \cdot {2}) + ({5}i \cdot {3}i) $ Then simplify the terms: $ (8) + (12i) + (10i) + (15 \cdot i^2) $ Imaginary unit multiples can be grouped together. $ 8 + (12 + 10)i + 15i^2 $ After we plug in $i^2 = -1$ , the result becomes $ 8 + (12 + 10)i - 15 $ The result is simplified: $ (8 - 15) + (22i) = -7+22i $